Question on double sum of proof of Countable subadditivity of outer measure I am reading Measure Theory from Sheldon Axler's Measure, Integration & Real Analysis. Here's the question that I have:
The rearrangement of $\{ I_{j,k} : j,k \in \mathbb{Z}^{+} \}$ (see proof below) is a little confusing to me. I understand that intuitively what the proof is doing is that we know that there is a function $\sigma : \mathbb{N} \to \mathbb{N}\times \mathbb{N}$ such that $\sigma (1)=(1,1)$, $\sigma(2)=(1,2)$, $\sigma(3)=(2,1)$ and so on and then we are claiming that
$$\sum \limits_{k=1}^{\infty} \mathscr{l}(I_{\sigma (k)})=\sum \limits_{k=1}^{\infty} \sum \limits_{j=1}^{\infty} \mathscr{l}(I_{j,k})$$
However I do not see how to prove this. I am not quite sure how such an rearrangement is possible. Hints will be appreciated in direction to prove this.

 A: The manipulation being done here is an application of the following:

Claim: If $a_{j,k}\ge 0$ for every $j,k=1,2,\ldots$, then $$\sum_{k=1}^\infty\sum_{j=1}^\infty a_{j,k}=\sum_{i=2}^\infty\sum_{j+k=i}a_{j,k}.$$

where $\sum_{j+k=i}a_{j,k}$ is shorthand for $a_{1,i-1}+a_{2,i-2}+\cdots+a_{i-1, 1}$.
One way to prove this rigorously is to first argue for each $j$, $k$:
$$a_{j,k}=\sum_{i=2}^\infty a_{j,k}[j+k=i]\tag1$$ where the Iverson bracket $[j+k=i]$ equals $1$ when $j+k=i$, and zero otherwise. (The reason why (2) is true is that every summand is zero except when $i=j+k$.) Now you can write
$$\sum_{k=1}^\infty\sum_{j=1}^\infty a_{j,k}=\sum_{k=1}^\infty\sum_{j=1}^\infty \sum_{i=2}^\infty a_{j,k}[j+k=i].\tag2$$
Since the summands on the RHS of (3) are all non-negative, it is permissible to interchange the order of summation. Do this repeatedly to obtain
$$\sum_{i=2}^\infty\left(\sum_{j=1}^\infty\sum_{k=1}^\infty a_{j,k}[j+k=i]\right)\tag 3.
$$
Finally argue that for each $i$ the quantity in parentheses reduces to a finite double sum over $j$ and $k$ which can be stopped at $i$: $$\sum_{j=1}^i\sum_{k=1}^i a_{j,k}[j+k=i]\tag4$$ This last is equal to $\sum_{j+k=i} a_{j,k}$, by simple enumeration.
(The claim and proof are a bit tidier if we index $j$ and $k$ starting from zero.)
